Hi all,

I'm busy on paper a lot lately and found a basic problem that I'm not sure off if I understand it correctly.

suppose we are on the turn and trying to find out if a pot sized bluffraise will be profitable to a pot sized bet.

We know we have about 30% equity when called.

The pot = $1.
And his bet = $1.
And we raise to $4.

so our ev = (1-call)*$2 + call*(0.3-0.5)*$9

For an EV greater than 0 we need him to call less than:

c < 2 / 3.8 -> c < 53%


Easy schmeasy... We can get away with a lot more calls than when we had a pure bluff here, because even if we're called we have some equity and therefore expect to win some money on showdown. This money is equity from the existing pot AND the bets/raises put in on this street.

The problem is that we had a street before this street, where we made a play just like this. We we're just breakeven on the play, because we had some equity there as well. But without that equity it would have been a losing play.

Now on this street we make up for our direct losses (compared to a pure bluff) with the equity from the bets and raises on this street, but also from the pot that was made last street. So this money from the pot is counted double in the overall strategy! It made up for some of the direct loss of the first play and we can't use it to make up for this play...

The solution is to change our EV into:

(1-call)*pot + call*(equity-0.5)*{money that went in on THIS street}

which equals: average win + average cost of the play itself (negative).



Am I right in this analysis? I have the feeling that in most poker math all the streets are looked at independently and the pot is noone's, but draw a simple tree from the story above and you'll see that two winning/breakeven plays after eachother can lead to a loosing strategy overall, because that 'promised money from our equity' is counted double.

There is another game theory thing that is bugging me...

We always compare our EV to folding on this street which we say is 0. To play somewhere close to optimal we want to keep our range as wide as possible. That means that if we can bet, raise or call with a higher ev than 0 we should do so (or we are giving up value). This also means that we chose our betting and raising ranges wide enough to be 0EV against the optimal counter strategy if we are playing optimally...

But... (without the rake) poker is a zero sum game. If we tailor our average EV to be 0 for any line, our opponents EV equals the pot, by definition. The money most go somewhere. So overall we will be losing all the money that we have put in on earlyer streets as soon as we tailor our strategy on that street to be 0EV.

I think the EV in an optimal strategy should be 0,5*pot, on average, on any street. This way both players play optimally and break even.

But then we have a problem... What do we do if all possible lines from a certain point have an EV smaller than 0,5pot?

Also how can we be indifferent between our actions if they are not all equal to a fold which is always 0EV? If we are not indifferent, how can our opponents strategy be optimal?

[QUOTE=mvdgaag;4884503]our ev = (1-call)*$2 + call*(0.3-0.5)*$9
QUOTE]

sorry. had a late night session... why is the call EV = call*(0.3-0.5)*$9? Shouldn't it be call EV = call * (0.3-0.7)*$9? Basically, if call = 1, you win 30% of the time and you lose 70% of the time. God! I can't think about this now - I need to get some sleep. :-)

I'm looking at overall strategies. So the EV should be compared to preflop.

So the pot = 0
We both ante 0,5
The pot is now 1
You bet 1
I call 1 and raise another 3


I win 30% of the time, so 70% of the time my stack is 4.50 lower compared to my starting stack, 30% of time time it is 4.50 higher.


0.3 * 4.50 - (1 - 0.3) * 4.50 = (2 * 0.3 - 1) * 4.50

or in general terms: (2*equity-1)*(1/2 pot)

which is the same as (equity - 0,5) * pot

Semibluffing is a positive EV strategy when your already behind in the hand...

Thus, if you analyze payoffs from preflop, by always assuming we end up in a semibluff situation, then yes our payoffs will (usually) be less than zero. I say usually, because it depends precisely on the type of semibluff situation u presume we end up in.

But semibluffing is a positive EV strategy once we see the flop (if properly combined with value raises, and a rational opponent capable of optimal folding). In otherwords, had we not semibluffed we'd have an even more negative ex-ante-preflop payoff than we would otherwise.

The problem with analyzing from preflop is that you can't always assume we end up in a semibluff situation. Also, even if you made that assumption, there are a few semibluff situations (albeit - non realistic) where your opponent can't call - and hence are awlways positive. (E.g. 3-flush, non-paired board of hearts, infinite stacks, and you Ah2s - u move all in with infinite stack - opponent must always fold. - u know that he cannot hold the nuts, he cannot know that you don't, and his pot odds are 1 to infinity). Even if your holding the semibluff hand more than 50% of the time he still loses because you have a 20% chance of drawing to the nuts anyways. Yes stupid example i admit, but i'll leave it to clever minds to formulate more realistic ones.

Finally, as for your nash equillibrium query - why would we use nash if it just makes everythign zero EV???... Its perhaps a confusion that so may explinations of Nash Equillibria involve the game rock paper scissors that people think nash equillibrium is above breaking even with every other strategy. Yes, lots of hedging is involved in Nash theory, often leaving your opponnent indifferent between calling and folding. But that doesn't mean we break even vs. all strategies using Nash.

Here's some mistakes you can make that a Nash strategy would exploit: Too loose / Too tight preflop range, too passive, too aggro - (bluffing with proper hands against nash ranges from zero to plus EV - but betting with too wide a range of value hands in negative EV), on a technical level failing to properly distribute bluff space, and many many more. Nash exploits these mistakes, and at other times it may give an opponent no room for mistake (e.g. an opponent deciding between calling, folding , the Nash strategy may have hedged our opponent into indifference).

Even though no one has calculated a perfect nash strategy, we know enough about what it looks like - and given the list of exploitable mistakes i already mentioned it should be obvious that no human could be able to do the calculations necessary to play break even against an opponent playing a perfect nash.

Thanks for the elaborate reply _D&L_!

I'll try to reply to your stuff and post all my headspam in between so you might get a sense of where I'm going...



Yes



I think we can... Anytime you bet or raise you have a known hand and an estimated range of your opponent. You can find the equity of your hand against his range. This will always we somewhere between 0% and 100%, but you will always have a given equity.

Now you will win the pot when your opponent folds and you will get your equity when called (this last assumption defines equity as: how often do we get to showdown times the percentage of wins we have when we get there)

Therefore any bet or raise is essentially a semibluff, even with the nuts or no outs... We talk about bluffs when we have almost no equity, semibluffs when we have decent equity, but need our opponent to fold a good amount of the time to break even and valuebets when we expect to win without our opponent folding a lot of his range if anything at all, but essentially it's all the same. Maybe I shouldn't call it a semibluff ...



No but we break even at worst, don't we? Although we will not be able to solve TH very soon I think it's possible and important to get a feel for where the tresholds roughly lie to understand where we deviate from them and how to exploit these deviations. I think the value of the postflop game cannot be much more than your opponents blind though, because if it would he'd be better off not playing.

That I want to break even on every street is because I believe we want to take every opportunity to give our opponent a chance to fold and give up the pot without giving him a chance to get more value from our weak range at showdown that he has lost by folding. If we play tighter than this we might win more at showdown, but we win less overall because we miss out on stealing opportinuties. If we play looser than this we will win a little more from our stealing, but lose a lot more from our weaker range when we get to showdown. I think an optimal strategy is perfectly in between and is the one that has the widest range going to showdown agressively that will not lose more at showdown than it wins from our opponents folds even if our opponent plays perfectly. In search of those equilibrium points I have done some simple calculations (on 0..1 and nuts/nothing games) and found out that breaking even on every street makes your overall game a losing one (and posted the OP).

So we really should keep our opponent indifferent between calling and folding (leaving our raising) at all streets and when we get to the river he should still be indifferent and win at most 33% of the time to a pot sized bet. This means he can call down lighter than we can bet, since we need 66% equity on the river to make him indifferent! This sounds very counter intuitive to me.... Don't you need more of a hand to call than to bet, in general? Gap principle?

I don't think its double counting to count the money put in on street 1, on street2. If your first bet has EV(0), and your second bet including that money has EV(0), then the sum of the two bluffs is still EV(0).

example

Lets play a game that isn't poker.
Lets say on betting round 1 I'm holding $1. I say to you, if you bet me $1, i'll give my $1 if this coin lands on heads, but you give me yours if it lands on tails.

EV of this betting round is zero (1/2 win 1$, 1/2 lose $1).

If you win, i go bankrupt and no future betting.
If I win though, have the money from the pot I just won, and lets assume u have another $2 in your pocket.

I say to you, ok this time I'll let u wager me $2 for my $2 and we'll flip the coin again.
Again zero EV() payoff.

This is essentially the same as the two street bluff u were hypothesizing in your original post. And I don't see two EV(0) bets becoming negative due to double counting.

1/2 chance u won coinf flip: EV = .50 (.50*1)
1/2 * 1/2 u lost first bet, but won second: EV = 0.25 (.25*(2-1))
1/2 * 1/2 u lost first bet, and lost second: EV = -.75 (.25*(-2-1))

Your coinflop example is correct.

But at the second flip I'll have 50% chance. In poker, when I want to bet I just need more EV than a fold. My EV of folding is 0. My EV of betting is 0 when I have only 33% equity, not 50%, because of the existing pot that we both created.

In your example in the second bet (given there is a pot and it's potsized) this would be 33%, not 50% as the coin gives us even is he calls 100% of the time.

EV call = 0.33 * 2 - 0.66 * 1 = 0

I was looking at the minimum equity we need to break even on a certain street when combined with folding equity.

We might breakeven on the second street, but on the first street we assumed that some of our investment would we won back (our equity). If we choose to breakeven on/from the next street we incorrectly made this assumption.